graph each equation. 9) \\(\\frac{x^2}{4} + \\frac{y^2}{9} = 1\\) graph with x from -8 to 8 and y from -8 to…

Explanation:

Step1: Identify the conic section

The equation (\frac{x^{2}}{4}+\frac{y^{2}}{9} = 1) is in the standard form of an ellipse (\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1) (where (a>b>0)), which is a vertical ellipse centered at the origin ((0,0)).

Step2: Find the values of (a) and (b)

For the equation (\frac{x^{2}}{4}+\frac{y^{2}}{9}=1), we can rewrite it as (\frac{x^{2}}{2^{2}}+\frac{y^{2}}{3^{2}} = 1). So, (a = 3) (semi - major axis length along the (y) - axis) and (b=2) (semi - minor axis length along the (x) - axis).

Step3: Find the vertices and co - vertices

• Vertices: The vertices of a vertical ellipse centered at the origin are ((0,\pm a)). Substituting (a = 3), the vertices are ((0,3)) and ((0, - 3)).
• Co - vertices: The co - vertices of a vertical ellipse centered at the origin are ((\pm b,0)). Substituting (b = 2), the co - vertices are ((2,0)) and ((- 2,0)).

Step4: Plot the points and draw the ellipse

1. Plot the vertices ((0,3)) and ((0,-3)) on the (y) - axis.
2. Plot the co - vertices ((2,0)) and ((-2,0)) on the (x) - axis.
3. Then, sketch the ellipse by connecting these points smoothly, making sure that the ellipse is symmetric about both the (x) - axis and (y) - axis.

(Note: Since the problem asks to graph the equation, the final answer is the graph of the ellipse with vertices at ((0, \pm3)) and co - vertices at ((\pm2,0)) centered at the origin, as described in the steps above. If we were to describe the key points for graphing: the ellipse is centered at ((0,0)), goes up to (y = 3), down to (y=-3), right to (x = 2), and left to (x=-2), and is symmetric about both axes.)

Answer:

The graph is an ellipse centered at the origin ((0,0)) with vertices ((0, 3)), ((0, - 3)) and co - vertices ((2, 0)), ((-2, 0)), sketched by connecting these points smoothly and symmetrically about the (x) - axis and (y) - axis.